\(\int \frac {(c+d x)^{5/4}}{(a+b x)^{7/2}} \, dx\) [449]

Optimal result

Integrand size = 19, antiderivative size = 175 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{7/2}} \, dx=-\frac {d \sqrt [4]{c+d x}}{3 b^2 (a+b x)^{3/2}}-\frac {d^2 \sqrt [4]{c+d x}}{6 b^2 (b c-a d) \sqrt {a+b x}}-\frac {2 (c+d x)^{5/4}}{5 b (a+b x)^{5/2}}-\frac {d^2 \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{6 b^{9/4} (b c-a d)^{3/4} \sqrt {a+b x}} \] Output:

-1/3*d*(d*x+c)^(1/4)/b^2/(b*x+a)^(3/2)-1/6*d^2*(d*x+c)^(1/4)/b^2/(-a*d+b*c 
)/(b*x+a)^(1/2)-2/5*(d*x+c)^(5/4)/b/(b*x+a)^(5/2)-1/6*d^2*(-d*(b*x+a)/(-a* 
d+b*c))^(1/2)*EllipticF(b^(1/4)*(d*x+c)^(1/4)/(-a*d+b*c)^(1/4),I)/b^(9/4)/ 
(-a*d+b*c)^(3/4)/(b*x+a)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.42 \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{7/2}} \, dx=-\frac {2 (c+d x)^{5/4} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-\frac {5}{4},-\frac {3}{2},\frac {d (a+b x)}{-b c+a d}\right )}{5 b (a+b x)^{5/2} \left (\frac {b (c+d x)}{b c-a d}\right )^{5/4}} \] Input:

Integrate[(c + d*x)^(5/4)/(a + b*x)^(7/2),x]
 

Output:

(-2*(c + d*x)^(5/4)*Hypergeometric2F1[-5/2, -5/4, -3/2, (d*(a + b*x))/(-(b 
*c) + a*d)])/(5*b*(a + b*x)^(5/2)*((b*(c + d*x))/(b*c - a*d))^(5/4))
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {57, 57, 61, 73, 765, 762}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(c+d x)^{5/4}}{(a+b x)^{7/2}} \, dx\)

\(\Big \downarrow \) 57

\(\displaystyle \frac {d \int \frac {\sqrt [4]{c+d x}}{(a+b x)^{5/2}}dx}{2 b}-\frac {2 (c+d x)^{5/4}}{5 b (a+b x)^{5/2}}\)

\(\Big \downarrow \) 57

\(\displaystyle \frac {d \left (\frac {d \int \frac {1}{(a+b x)^{3/2} (c+d x)^{3/4}}dx}{6 b}-\frac {2 \sqrt [4]{c+d x}}{3 b (a+b x)^{3/2}}\right )}{2 b}-\frac {2 (c+d x)^{5/4}}{5 b (a+b x)^{5/2}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {d \left (\frac {d \left (-\frac {d \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}}dx}{2 (b c-a d)}-\frac {2 \sqrt [4]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{6 b}-\frac {2 \sqrt [4]{c+d x}}{3 b (a+b x)^{3/2}}\right )}{2 b}-\frac {2 (c+d x)^{5/4}}{5 b (a+b x)^{5/2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {d \left (\frac {d \left (-\frac {2 \int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [4]{c+d x}}{b c-a d}-\frac {2 \sqrt [4]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{6 b}-\frac {2 \sqrt [4]{c+d x}}{3 b (a+b x)^{3/2}}\right )}{2 b}-\frac {2 (c+d x)^{5/4}}{5 b (a+b x)^{5/2}}\)

\(\Big \downarrow \) 765

\(\displaystyle \frac {d \left (\frac {d \left (-\frac {2 \sqrt {1-\frac {b (c+d x)}{b c-a d}} \int \frac {1}{\sqrt {1-\frac {b (c+d x)}{b c-a d}}}d\sqrt [4]{c+d x}}{(b c-a d) \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}-\frac {2 \sqrt [4]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{6 b}-\frac {2 \sqrt [4]{c+d x}}{3 b (a+b x)^{3/2}}\right )}{2 b}-\frac {2 (c+d x)^{5/4}}{5 b (a+b x)^{5/2}}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {d \left (\frac {d \left (-\frac {2 \sqrt {1-\frac {b (c+d x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{\sqrt [4]{b} (b c-a d)^{3/4} \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}-\frac {2 \sqrt [4]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{6 b}-\frac {2 \sqrt [4]{c+d x}}{3 b (a+b x)^{3/2}}\right )}{2 b}-\frac {2 (c+d x)^{5/4}}{5 b (a+b x)^{5/2}}\)

Input:

Int[(c + d*x)^(5/4)/(a + b*x)^(7/2),x]
 

Output:

(-2*(c + d*x)^(5/4))/(5*b*(a + b*x)^(5/2)) + (d*((-2*(c + d*x)^(1/4))/(3*b 
*(a + b*x)^(3/2)) + (d*((-2*(c + d*x)^(1/4))/((b*c - a*d)*Sqrt[a + b*x]) - 
 (2*Sqrt[1 - (b*(c + d*x))/(b*c - a*d)]*EllipticF[ArcSin[(b^(1/4)*(c + d*x 
)^(1/4))/(b*c - a*d)^(1/4)], -1])/(b^(1/4)*(b*c - a*d)^(3/4)*Sqrt[a - (b*c 
)/d + (b*(c + d*x))/d])))/(6*b)))/(2*b)
 

Defintions of rubi rules used

rule 57

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & 
& GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m 
+ n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c 
, d, m, n, x]
 

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 762

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) 
)*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] 
 && GtQ[a, 0]
 

rule 765

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[Sqrt[1 + b*(x^4/a)]/Sqrt 
[a + b*x^4]   Int[1/Sqrt[1 + b*(x^4/a)], x], x] /; FreeQ[{a, b}, x] && NegQ 
[b/a] &&  !GtQ[a, 0]
 

Maple [F]

\[\int \frac {\left (x d +c \right )^{\frac {5}{4}}}{\left (b x +a \right )^{\frac {7}{2}}}d x\]

Input:

int((d*x+c)^(5/4)/(b*x+a)^(7/2),x)
 

Output:

int((d*x+c)^(5/4)/(b*x+a)^(7/2),x)
 

Fricas [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{7/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((d*x+c)^(5/4)/(b*x+a)^(7/2),x, algorithm="fricas")
 

Output:

integral(sqrt(b*x + a)*(d*x + c)^(5/4)/(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2* 
x^2 + 4*a^3*b*x + a^4), x)
 

Sympy [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{7/2}} \, dx=\int \frac {\left (c + d x\right )^{\frac {5}{4}}}{\left (a + b x\right )^{\frac {7}{2}}}\, dx \] Input:

integrate((d*x+c)**(5/4)/(b*x+a)**(7/2),x)
 

Output:

Integral((c + d*x)**(5/4)/(a + b*x)**(7/2), x)
 

Maxima [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{7/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((d*x+c)^(5/4)/(b*x+a)^(7/2),x, algorithm="maxima")
 

Output:

integrate((d*x + c)^(5/4)/(b*x + a)^(7/2), x)
 

Giac [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{7/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {5}{4}}}{{\left (b x + a\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((d*x+c)^(5/4)/(b*x+a)^(7/2),x, algorithm="giac")
 

Output:

integrate((d*x + c)^(5/4)/(b*x + a)^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{7/2}} \, dx=\int \frac {{\left (c+d\,x\right )}^{5/4}}{{\left (a+b\,x\right )}^{7/2}} \,d x \] Input:

int((c + d*x)^(5/4)/(a + b*x)^(7/2),x)
 

Output:

int((c + d*x)^(5/4)/(a + b*x)^(7/2), x)
 

Reduce [F]

\[ \int \frac {(c+d x)^{5/4}}{(a+b x)^{7/2}} \, dx=\text {too large to display} \] Input:

int((d*x+c)^(5/4)/(b*x+a)^(7/2),x)
 

Output:

(16*(c + d*x)**(1/4)*sqrt(a + b*x)*a*c*d - 4*(c + d*x)**(1/4)*sqrt(a + b*x 
)*a*d**2*x + 20*(c + d*x)**(1/4)*sqrt(a + b*x)*b*c**2 + 40*(c + d*x)**(1/4 
)*sqrt(a + b*x)*b*c*d*x + 5*int(((c + d*x)**(1/4)*sqrt(a + b*x)*x)/(a**5*c 
*d + a**5*d**2*x - 10*a**4*b*c**2 - 6*a**4*b*c*d*x + 4*a**4*b*d**2*x**2 - 
40*a**3*b**2*c**2*x - 34*a**3*b**2*c*d*x**2 + 6*a**3*b**2*d**2*x**3 - 60*a 
**2*b**3*c**2*x**2 - 56*a**2*b**3*c*d*x**3 + 4*a**2*b**3*d**2*x**4 - 40*a* 
b**4*c**2*x**3 - 39*a*b**4*c*d*x**4 + a*b**4*d**2*x**5 - 10*b**5*c**2*x**4 
 - 10*b**5*c*d*x**5),x)*a**6*d**4 - 60*int(((c + d*x)**(1/4)*sqrt(a + b*x) 
*x)/(a**5*c*d + a**5*d**2*x - 10*a**4*b*c**2 - 6*a**4*b*c*d*x + 4*a**4*b*d 
**2*x**2 - 40*a**3*b**2*c**2*x - 34*a**3*b**2*c*d*x**2 + 6*a**3*b**2*d**2* 
x**3 - 60*a**2*b**3*c**2*x**2 - 56*a**2*b**3*c*d*x**3 + 4*a**2*b**3*d**2*x 
**4 - 40*a*b**4*c**2*x**3 - 39*a*b**4*c*d*x**4 + a*b**4*d**2*x**5 - 10*b** 
5*c**2*x**4 - 10*b**5*c*d*x**5),x)*a**5*b*c*d**3 + 15*int(((c + d*x)**(1/4 
)*sqrt(a + b*x)*x)/(a**5*c*d + a**5*d**2*x - 10*a**4*b*c**2 - 6*a**4*b*c*d 
*x + 4*a**4*b*d**2*x**2 - 40*a**3*b**2*c**2*x - 34*a**3*b**2*c*d*x**2 + 6* 
a**3*b**2*d**2*x**3 - 60*a**2*b**3*c**2*x**2 - 56*a**2*b**3*c*d*x**3 + 4*a 
**2*b**3*d**2*x**4 - 40*a*b**4*c**2*x**3 - 39*a*b**4*c*d*x**4 + a*b**4*d** 
2*x**5 - 10*b**5*c**2*x**4 - 10*b**5*c*d*x**5),x)*a**5*b*d**4*x + 105*int( 
((c + d*x)**(1/4)*sqrt(a + b*x)*x)/(a**5*c*d + a**5*d**2*x - 10*a**4*b*c** 
2 - 6*a**4*b*c*d*x + 4*a**4*b*d**2*x**2 - 40*a**3*b**2*c**2*x - 34*a**3...